Question

a 6.00-kg ornament is held at rest by two light wires that form 30° angles with the vertical, as shown in the figure. an external force of magnitude f acts vertically downward on the ornament. the tension exerted by each of the two wires is denoted by t. a free-body diagram, showing the four forces that act on the box, is shown in the figure. if the magnitude of force f is 410 n, what is the magnitude of the tension t?

Answer 1

The equilibrium condition allows finding the result for the tension of the cable that supports the block is:

• The tension is:  T₁ = 270.66 N

Newton's second law gives a relationship between force, mass and the acceleration of bodies, when the acceleration is zero it is called the equilibrium condition.

A free-body diagram of the forces is shown in the attachment.

Let's write the equilibrium equation for the vertical axis.

        2 $T_1_y$ - f - W = 0

          2 $T_1_y$  = W + f

Let's use trigonometry to find the tension, as the angle indicates that it is relative to the vertical.

        cos 30 = $\frac{T_1_y}{T_1}$  

        $T_1_y$  = T1 cos 30

     

We substitute.

        2T₁ cos 30 = W + f

         T₁ = m g + f / 2 cos 30

         T₁ = $\frac{6 \ 9.8 + 410}{2 \ cos 30}$  

         T₁ = 270.66 N

In conclusion using the equilibrium condition we can find the result for the tension of the cable that supports en bloc is:

•   The tension is:  270.66 N

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